Sign-Change Methods (OCR GCSE Maths)

Revision Note

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Mark

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Mark

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Maths

Sign-Change Methods

You must know either the Decimal Search method or the Interval Bisection method (exam questions will not force you to use a particular method)

What is iteration?

  • Iteration means finding an approximation to the solution using a method that gets closer and closer (converges) to the real solution
  • It is often used for equations which don't have nice solutions
    • solutions themselves might be irrational numbers

How do you show that there is a solution between two values?

  • To find x0 (the initial / starting value), you are often asked to show that there is a solution between two values
  • For example, show that there is a solution to x3 + x = 7 between 1 and 2
    • Method 1: Leave a constant term (e.g. the 7) on the right, substitute x = 1 and x = 2 into the left and show that this gives values below and above 7
      • 13 + 1 = 2 and 23 + 2 = 10 which are below and above 7
      • A solution therefore lies between 1 and 2
    • Method 2: Use "0" as your constant term on the right (by rearranging the equation into "... = 0"), then substitute in x = 1 and x = 2, showing this gives values below and above 0, i.e. negative and positive
    • this is called a change of sign between 1 and 2
      • x3 + x - 7 = 0
      • Substitute x = 1 into the left-hand side: 13 + 1 - 7 = -5 (negative)
      • Substitute x = 2 into the left-hand side:  23 + 2 - 7 = 3 (positive)
      • A solution lies between 1 and 2 as there is a change of sign

Decimal Search

  • Let’s say you know that a solution to an equation lies between 1 and 2 and you want to find it to 1 decimal place
    • The equation must be rearranged to have “= 0” on one side
  • Search every decimal, by substituting 1.0, 1.1, 1.2, 1.3, 1.4, … into the equation until you see a change of sign
    • Let’s say there was a change of sign between 1.6 and 1.7
    • This is not yet enough information to be able to round the solution to 1 decimal place
  • Substitute in 1.65, the value exactly in the middle of 1.6 and 1.7
    • If the sign change is between 1.6 and 1.65, the solution rounds down to 1.6 to 1 decimal place
    • If the sign change is between 1.65 and 1.7, the solution rounds up to 1.7 to 1 decimal place

Interval Bisection

  • Let’s say you know that a solution to an equation lies between 1 and 2 and you want to find it to 1 decimal place
    • The equation must be rearranged to have “= 0” on one side
  • Bisect (find the mid-point of) the interval from 1 to 2 to get 1.5
    • Substitute 1, 1.5 and 2 into the equation and see where the sign change is
  • Let’s say the sign change is between 1.5 and 2
    • Bisect this interval to get 1.75 and substitute this value in
  • Let’s say the sign change is now between 1.5 and 1.75
    • Bisect this interval to get 1.625 and substitute this value in
  • Keep going until your interval is small enough to give you only one option to round your solution to
    • For example, 1.65625 < x < 1.6875 means that the solution is 1.7 to 1 decimal place

Exam Tip

  • Always look at the accuracy asked for in the question to know how far to go
  • The decimal search method is more commonly used (it is the same idea as "trial and improvement")

Worked example

A solution to the equation x3 + x = 7 lies between 1 and 2.
Find this solution correct to 1 decimal place.
 

Method 1: Decimal Search
Rearrange the equation to have = 0 on one side

 

x3 + x - 7 = 0
 

Substitute 1.0, 1.1, 1.2, 1.3, 1.4, … into  the equation until a change of sign happens
 

x

x3 + x – 7

1.0

-5

1.1

-4.569

1.2

-4.072

1.3

-3.503

1.4

-2.856

1.5

-2.125

1.6

-1.304

1.7

-0.387

1.8

0.632

 

A change of sign is between 1.7 and 1.8 so test 1.75
 

When x = 1.75,  x3 + x – 7 = 0.109375
 

See if the change of sign is between 1.7 and 1.75 or 1.75 and 1.8
 

The change of sign is between 1.7 and 1.75
 

The solution rounds to 1.7 to 1 decimal place

1.7 to 1 decimal place

Method 2: Interval Bisection
Rearrange the equation to have = 0 on one side
 

x3 + x - 7 = 0
 

1.5 bisects the interval 1 to 2
Substitute 1.0, 1.5 and 2.0 into the equation

 

x

x3 + x – 7

1.0

-5

1.5

-2.125

2.0

3


There is a change of sign between 1.5 and 2
Bisect this interval (1.75) and substitute this value into the equation

 


When x = 1.75,  x3 + x – 7 = 0.109375
 

There is a change of sign between 1.5 and 1.75
Bisect this interval, (1.5 + 1.75)
÷ 2 = 1.625, and substitute this value into the equation
 


When x = 1.625,  x3 + x – 7 = -1.084
 

There is a change of sign between 1.625 and 1.75
Bisect this interval, (1.625 + 1.75)
÷ 2 = 1.6875, and substitute this value into the equation
 


When x = 1.6875,  x3 + x – 7 = -0.507...
 

There is a change of sign between 1.6875 and 1.75
This is enough to know that the solution rounds to 1.7 (it cannot be 1.75 itself as that gave 0.109375 from the equation)

1.7 to 1 decimal place

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Mark

Author: Mark

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.